It takes $1 \, min$ for hot water to cool from $80^{\circ}C$ to $60^{\circ}C$. Find the time taken (in $sec$) to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the surroundings is $30^{\circ}C$.

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120^{\circ}C$,then

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

$A$ solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,

$A$ sphere of density $\rho$,specific heat capacity $c$,and radius $r$ is hung by a thermally insulating thread in an enclosure which is kept at a lower temperature than the sphere. The temperature of the sphere starts to drop at a rate which depends upon the temperature difference between the sphere and the enclosure and the nature of the surface of the sphere and is proportional to

State Newton's law of cooling and derive its mathematical equation.